Problem: In triangle $ABC$, $AB=AC$ and $D$ is a point on $\overline{AC}$ so that $\overline{BD}$ bisects angle $ABC$. If $BD=BC$, what is the measure, in degrees, of angle $A$?
Answer: Since $AB=AC$, triangle $ABC$ must be an isosceles triangle and the measures of $\angle ABC$ and $\angle ACB$ must be equal. Continuing, since $\overline{BD}$ bisects angle $ABC$, we have that the measures of $\angle ABD$ and $\angle BDC$ are equal. Finally, since $BD=BC$, triangle $BDC$ must also be an isosceles triangle so the measures of $\angle BDC = \angle BCD$. Now if we consider triangle $BDC$, we know that angles $BDC$ and $BCD$ have equal angle measures and angle $DBC$ has an angle measure that is half that of the other two. Since these three angle measures must add up to $180^\circ$, we have that $\angle DBC$ has measure $36^\circ$ and angles $BDC$ and $BCD$ have measures $72 ^\circ$.

Now, since $\angle ABC \cong \angle ACB$ and $\angle ACB$ has measure $72^\circ$, we know that $\angle A$ must have an angle measure of $180-72-72=\boxed{36}$ degrees.